Thursday, January 5, 2012

Problem 714: Triangle, Parallel, Cevians, Midpoints, Median, Ceva Theorem

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 714 details.

Online Geometry Problem 714: Triangle, Parallel, Cevians, Midpoints, Median, Ceva Theorem.

4 comments:

  1. Minor correction due to typo error:
    Apply Ceva's theorem in tri. ABC and cevians AE,CD and BG.
    (DA/DB).(EB/EC).(GC/GA)=1 ----(1)
    Since DE//AC >>>DA/DB=EC/EB
    Replace this in (1) and simplify we get GC/GA=1 >> G is the midpoint of AC.
    Using similar triangles with // lines we have
    DH/AG=HE/GC=BH/BG >> HD=HE >> H is the midpoint of DE

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  2. Draw AM ⊥ BG(extended)and CN ⊥ BG.
    Denote area by (...)
    (ADC)=(AEC)(enclosed between parallels and having common base)
    Remove common area (AFC) from either side:
    (ADF) = (CEF)
    Add common area (DBEF) to either side:
    (BDC)=(BEA)
    They have a common base BF
    So AM = CN. Also AM ∥ CN
    ∆AMG ≡ ∆CNG (a.s.a)
    ∴AG = GC
    AG being median bisects DE too (∵ DE ∥ AC)

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  3. To Pravin:( refer to your solution)
    area of (BDC)= (BEA) is correct but these 2 triangles do not have common base.
    Please explain.

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  4. Revised and corected Solution of Problem 714 (Thanks to Tran):

    Draw AM ⊥ BG(extended)and CN ⊥ BG.
    Denote area by (...)
    (ADC)=(AEC)(enclosed between parallels and having common base)
    Remove common area (AFC) from either side:
    (ADF) = (CEF)
    Add common area (DBEF) to either side:
    (BDC)=(BEA) [as proved earlier]

    Now,
    AG/GC = (ABG)/(CBG) = (AFG)/(CFG)
    = Diff of Nrs / Diff of Drs
    = (ABF)/(CBF) --->(1)

    By similar triangles AG/HE = FG/HF = GC/DH
    ∴ AG/GC = HE/DH --->(2)

    Next HE/DH = (BHE)/(BHD) = (HFE)/(HFD)
    =Sum of Nrs./Sum of Drs.= (BFE)/(BFD)--->(3)

    From (1), (2), (3)
    AG/GC = HE/DH = (ABF)/(CBF) = (BFE)/(BFD)
    = sum of Nrs./Sum of Drs.= (ABE)/(BCD) =1

    Hence AG = GC and DH = HE

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